Math
posted by Jake .
Find the value of the constant t such that the triangle formed by A (2,2) ,B (4,5) ,C (6,t) is right angled at B(4,5).

Line passing through two given points P1(x1,y1) and P2(x2,y2)
L: (yy1)/(y2y1) = (xx1)/(x2x1)
or (x2x1)(yy1) = (y2y1)(xx1)
For line AB:
AB: (42)(y2) = (52)(x2)
=> AB: 2(y2) = (7)(x2)
=> AB: y = 3.5x + 9
Similarly, line BC is:
BC: (64)*(y+5)=(t+5)*(x4)
=> BC: y=((t+5)/2)x  2t15
If the two lines are to be perpendicular to each other, the product of the slopes must be 1, or
3.5*(t+5)/2 = 1
Solve for t to get t=31/7
Check:
Slope of AB: (52)/(42)=3.5
Slope of BC: (31/7+5)/(64)= 2/7
Product of slopes : 3.5*2/7=1 OK
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